Question

Evaluate the integrals in Exercises 33–52.∫ from -π/4 to π/4 of 6 tan⁴(x) dx

Accepted Answer

Recognize that the integral is $$ \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} 6 	an^{4}(x) \, dx . $$The constant 6 can be factored out of the integral, so rewrite it as $$ 6 \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} 	an^{4}(x) \, dx . $$Note that $$ 	an^{4}(x)  is an $$even function because $$ 	an(-x) = -	an(x) $$, so $$ 	an^{4}(-x) = 	an^{4}(x) . $$This means the integral from $$ -\frac{\pi}{4}  to$$ $$ \frac{\pi}{4} $$ can be simplified to twice the integral from 0 to $$ \frac{\pi}{4} $$: $$ 6 	imes 2 \int_0^{\frac{\pi}{4}} 	an^{4}(x) \, dx = 12 \int_0^{\frac{\pi}{4}} 	an^{4}(x) \, dx . $$Express $$ 	an^{4}(x)  in $$terms of $$ \sec(x) $$ and $$ 	an(x) $$ using the identity $$ 	an^{2}(x) = \sec^{2}(x) - 1 . So$$, $$ 	an^{4}(x) = (	an^{2}(x))^{2} = (\sec^{2}(x) - 1)^{2} = \sec^{4}(x) - 2\sec^{2}(x) + 1 . $$Rewrite the integral as $$ 12 \int_0^{\frac{\pi}{4}} (\sec^{4}(x) - 2\sec^{2}(x) + 1) \, dx = 12 \left( \int_0^{\frac{\pi}{4}} \sec^{4}(x) \, dx - 2 \int_0^{\frac{\pi}{4}} \sec^{2}(x) \, dx + \int_0^{\frac{\pi}{4}} 1 \, dx ight) . $$Evaluate each integral separately: use reduction formulas or known integrals for $$ \int \sec^{2}(x) \, dx $$ and $$ \int \sec^{4}(x) \, dx $$, and the integral of 1 is straightforward. Then combine the results and multiply by 12 to express the final integral.

Evaluate the integrals in Exercises 33–52.
∫ from -π/4 to π/4 of 6 tan⁴(x) dx

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